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| Any chance for nomenclator? |
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Posted by: Anton - 25-02-2023, 12:36 AM - Forum: Analysis of the text
- Replies (18)
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Two arguments against the nomenclator cipher hypothesis that I have met with in discussions were:
1) It would be unusual to encipher a whole book in such manner
2) It would require a huge nomenclator to encipher a volume of this size
Considering argument #1, I think that the VMS is cryptologically unusual (if not unique) anyway, so the argument of unusualness-to-be does not look a strong one to me.
Considering argument #2, I tried to estimate the size of the nomenclator, assuming what is supposedly the worst-case scenario (OK, upon further consideration it is not the worst case of course, but let's say, a "worse case") - namely, that each unique plain text word is matched to a unique vord.
According to voynich.nu, the vocabulary of the VMS is comprised of ~8100 unique vords. That would be the size of the nomenclator in question. However, the nomenclator must contain not only vords, but also their translations into words. This way, the amount of tokens in the nomenclator doubles and results in 8100 x 2 = 16200.
What folio space would be required to hold that many of tokens? This we can deduce from Q20 which presents a convenient example of folio space filled almost exclusively with text, without drawings. 23 pages of Q20 (excluding the mostly empty f116v) contain circa 10700 vords. This means the average density of 10700/23 = 470 vords per page. Let's even assume a 30% margin for breaking text into columns, leaving more free space or increasing the letter size for better legibility etc. This yields the density of 470 x 0.7 = 330 vords per page. Assuming this density for the nomenclator space, 16200 tokens will occupy 49 pages or (regarding a "folio" in this case as a two-sided sheet) 49/2 = 25 folios. This is roughly two quires of the size of the present state of Q20 (without the missing folios). In the relative figures, this would be a (234 + 49)/234 = 21% increase in the total "thickness" of the MS.
The figures look significant but not like impossible.
This calculation implicitly assumes roughly equal average token length both for plain text and cipher text, but this would not be something unexpected judging by the average vord length.
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| A priori or A posteriori cracking? |
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Posted by: Arichichi - 23-02-2023, 03:46 PM - Forum: Analysis of the text
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Which is better for cracking the Voynich Manuscript? Thinking that a certain page talks about what is displayed in the page? Or thinking that it has no specific meaning relative to the featured pictures in the page? To rephrase my question, I want to know whether the cryptologists that deal with languages have ever tried to crack natural languages without a priori knowledge, just to test how strong their methods are. For example, has anyone tried to learn Korean or a similar language by studying a random Korean book and relying only on that book just to test the effectiveness of the code breaking methods that he relies on?
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Division of Voynich manuscript into 5 parts |
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Posted by: Addsamuels - 19-02-2023, 08:11 PM - Forum: Analysis of the text
- Replies (4)
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How do you categorise the voynich manuscript?
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René Z writes these categories.
Also I would like the folio numbers too.
Regards,
I have the Currier A and B dichotomy and You are not allowed to view links. Register or Login to view. helped greatly.
[PS: a lot of my textual analysis is useless and thus no results are being published]
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| Suggestions for EVA |
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Posted by: nablator - 13-02-2023, 12:14 PM - Forum: Analysis of the text
- Replies (8)
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Hi René,
I have an unfinished transliteration with some non-standard notations that I could (maybe) make compatible with some flavor of the EVA format for sharing.
I used:
, for uncertain spaces: when the spacing is noticeably larger than in the rest of the word but smaller than a half-space
; for half-spaces: they are not uncertain, large enough to insert a i or small e
. for full spaces (as large as the previous glyph or close)
.; for large spaces, not as large as double spaces
.. for double spaces (as large as two previous glyphs)
/ and \ for (large) vertical offsets with less than a full horizontal space
[e] for the rare e under a gallows leg
+ for the possibly fused glyphs (with a common part): a+r instead of a', a+n instead of u, etc.
";" conflicts with the HTML-like "@number;" or maybe not, because ";" is not really necessary after @number for parsing so I omitted it.
"[e]" conflicts with the "[:]" notation, or maybe not, because there is no ":".
Any suggestions? Thanks.
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| Folio 43r |
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Posted by: Ruby Novacna - 11-02-2023, 01:55 PM - Forum: Voynich Talk
- Replies (4)
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While continuing to search for words containing the combination pch, I came across the word cheol!keepchy/ cheol.keepchy, which is unique in the text.
Could this word be the name of the plant?
I read this word as kianuph9 and find it quite similar to the word κεάν-ωθος - corn-thistle, Carduus arvensis.
I have not yet looked up the identifications of the plant made earlier.
It is true that in this reading one vowel and one consonant are not identical, however it may be a dialect that would pronounce omega as upsilon and theta as f?
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| Archive Proximity |
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Posted by: Mark Knowles - 10-02-2023, 01:33 PM - Forum: Voynich Talk
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I have bemoaned the fact that Voynich research seems to largely ignore the wealth of material in archives. I appreciate the fact that depending on where ones lives actually visiting an archive may be difficult or inconvenient at best and so online searching is much more convenient. However what percentage of material is available online; if my cipher experience is something to go by then maybe 10%, not more than 20%, which leaves 80+% unexplored. Traveling to another country to do archival research is expensive and time consuming, so not possible for many people. If one is fortunate enough to live near a major archive then the situation is different.
I am fortunate to live near the Bodleian library in Oxford, which is large archive and contains material from continental Europe from the time of Voynich. However someone living in Rome or Paris or Milan amongst other cities would be better placed than I am to find relevant material. I get the impression that the number of researchers living in such locations is very few. Why is that? It would be intriguing to see a map highlighting the most relevant and significant archives to Voynich research and a map showing the cities where Voynich researchers live.
I myself am seriously considering moving to Italy some day in the future as there is a lot more that I can do there than here.
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| k/t gallows reduplication |
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Posted by: nablator - 09-02-2023, 10:05 PM - Forum: Analysis of the text
- Replies (22)
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I don't remember reading about how the k/t gallows sequence compares to random and human-generated pseudo-random sequences. It is likely that I missed something, so please point me to any existing analysis.
There is a blog post by Emma May Smith but it does not discuss the reduplication statistics: You are not allowed to view links. Register or Login to view.
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TLDR: The k/t gallows sequence is strongly biased toward reduplication; especially long sequences of the same k/t gallows in a row.
What is interesting about it is that human-generated pseudo-random sequences have a "tendency to overalternate between outcomes" documented in psychology studies. See for example: You are not allowed to view links. Register or Login to view. and the referenced literature.
Why do we see the opposite tendency in the VMS?
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EVA-k is more frequent than EVA-t. In the ZL transliteration, in paragraphs only, there are:
k = 10096, probability(k) = pk = 62%
t = 6063, probability(t) = pt = 38%
In few cases of ambiguity, I kept the first: for example [k:t] = k.
With perfectly random independent draws, in any window of n letters of the k/t sequence, the probability of having:
n times k is pk^n
n times t is pt^n
The expected number of windows of n identical letters in the k/t sequence is the probability multiplied by the number of windows:
For k: (k+t-n+1)*pk^n
For t: (k+t-n+1)*pt^n
The expected numbers are ek, et (rounded), the actual numbers are ak, at:
n ek < ak et < at
2 6307 6956 2274 2924
3 3940 4938 853 1554
4 2461 3593 320 870
5 1538 2686 120 501
6 960 2029 45 300
7 600 1565 16 184
8 375 1214 6 117
9 234 950 2 76
10 146 752 0 50
11 91 595 0 32
12 57 465 0 21
Note: if you count kk... and tt... sequences with a text editor like Notepad++, the numbers will be lower, because it skips to the text following the matched pattern instead of doing a "rolling window" search.
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